Solve the following quadratic equations. x^2-(5-i) x+(18+i)=0

Question

Solve the following quadratic equations.
$x^2-(5-i) x+(18+i)=0$

✓

Answer

Given equation is $x^2-(5-i) x+(18+i)=0$
Comparing with $a x^2+b x+c=0$, we get
$
a=1, b=-(5-i), c=18+i
$
Discriminant $=b^2-4 a c$
$
\begin{aligned}
& =[-(5-i)]^2-4 \times 1 \times(18+i) \\
& =25-10 i+i^2-72-4 i \\
& =25-10 i-1-72-4 i \ldots .\left[\div i^2=-1\right] \\
& =-48-14 i
\end{aligned}
$
So, the given equation has complex roots.
These roots are given by
$
\begin{aligned}
x & =\frac{-\mathrm{h} \pm \sqrt{\mathrm{b}^2-4 a \mathrm{c}}}{2 \mathrm{a}} \text { } \\
& =\frac{-[-(5-\mathrm{i})] \pm \sqrt{-48-14 \mathrm{i}}}{2(1)}=\frac{(5-\mathrm{i}) \pm \sqrt{-48-14 \mathrm{i}}}{2}
\end{aligned}
$
Let $\sqrt{-48-14 i}=a+b i$, where $a, b \in R$
Squaring on both sides, we get
$
\begin{aligned}
& -48-14 \mathrm{i}=\mathrm{a}^2+\mathrm{b}^2 \mathrm{i}^2+2 \mathrm{abi} \\
& -48-14 \mathrm{i}=\left(\mathrm{a}^2-\mathrm{b}^2\right)+2 \mathrm{abi} \quad \ldots\left[\because \mathrm{i}^2--1\right]
\end{aligned}
$
Equating real and imaginary parts, we get $a^2-b^2--48$ and $2 a b=-14$
$
\begin{array}{ll}
\therefore & a^2-b^2=-48 \text { and } b=\frac{-7}{a} \\
\therefore & a^2-\left(\frac{-7}{a}\right)^2=-48 \\
\therefore & a^2-\frac{49}{a^2}=-48 \\
\therefore & a^4-49--48 a^2 \\
\therefore & a^4+48 a^2-49=0 \\
\therefore & \left(a^2+49\right)\left(a^2-1\right)=0 \\
\therefore & a^2=-49 \text { or } a^2=1 \\
& \text { But } a \in \mathbb{R} \\
\therefore & a^2 \neq-49 \\
\therefore & a^2=1 \\
\therefore & a= \pm 1
\end{array}
$
When $a=1, b=\frac{7}{1}=-7$
When $a=-1, b=\frac{-7}{-1}=7$
$
\begin{array}{ll}
\therefore & \sqrt{-48-14 i}- \pm(1-7 i) \\
\therefore & x=\frac{(5-i) \pm(1-7 i)}{2} \\
\therefore & x=\frac{5-i+1-7 i}{2} \text { or } x=\frac{5-i-1+7 i}{2} \\
\therefore & x=3-4 i \text { or } x=2+3 i
\end{array}
$